How is the series 4+11+18+...+319 represented in summation notation?

7

Step-by-step explanation:

a1 =a1+(n-1)*d =4+(1-1)*7 =4

a2 =a1+(n-1)*d =4+(2-1)*7 =11

a3 =a1+(n-1)*d =4+(3-1)*7 =18

a4 =a1+(n-1)*d =4+(4-1)*7 =25

a5 =a1+(n-1)*d =4+(5-1)*7 =32

a6 =a1+(n-1)*d =4+(6-1)*7 =39

a7 =a1+(n-1)*d =4+(7-1)*7 =46

a8 =a1+(n-1)*d =4+(8-1)*7 =53

a9 =a1+(n-1)*d =4+(9-1)*7 =60

a10 =a1+(n-1)*d =4+(10-1)*7 =67

a11 =a1+(n-1)*d =4+(11-1)*7 =74

a12 =a1+(n-1)*d =4+(12-1)*7 =81

a13 =a1+(n-1)*d =4+(13-1)*7 =88

a14 =a1+(n-1)*d =4+(14-1)*7 =95

a15 =a1+(n-1)*d =4+(15-1)*7 =102

a16 =a1+(n-1)*d =4+(16-1)*7 =109

a17 =a1+(n-1)*d =4+(17-1)*7 =116

a18 =a1+(n-1)*d =4+(18-1)*7 =123

a19 =a1+(n-1)*d =4+(19-1)*7 =130

a20 =a1+(n-1)*d =4+(20-1)*7 =137

a21 =a1+(n-1)*d =4+(21-1)*7 =144

a22 =a1+(n-1)*d =4+(22-1)*7 =151

a23 =a1+(n-1)*d =4+(23-1)*7 =158

a24 =a1+(n-1)*d =4+(24-1)*7 =165

a25 =a1+(n-1)*d =4+(25-1)*7 =172

a26 =a1+(n-1)*d =4+(26-1)*7 =179

a27 =a1+(n-1)*d =4+(27-1)*7 =186

a28 =a1+(n-1)*d =4+(28-1)*7 =193

a29 =a1+(n-1)*d =4+(29-1)*7 =200

a30 =a1+(n-1)*d =4+(30-1)*7 =207

a31 =a1+(n-1)*d =4+(31-1)*7 =214

a32 =a1+(n-1)*d =4+(32-1)*7 =221

a33 =a1+(n-1)*d =4+(33-1)*7 =228

false

step-by-step explanation:

false : it can not go to 9 the highest that part can go is 7

here is your proof